# Power, impedance and math - did I do this right?

Discussion in 'Archived Threads 2001-2004' started by Patrick Sun, Feb 21, 2001.

1. ### Patrick Sun Moderator Moderator

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Okay, this may get a little scary with the math (not really), but here goes:
I'm working out equalizing my center channel components, which has a tweeter (8 ohm), midrange (8 ohm) and 2 woofers (both 8 ohm). I need the 2 woofer for symmetry.
The tweeter and the midrange have current divider networks of resistors to bring their SPL to around 87dB/1W/1m. What I want to do it make sure the woofer will give me the same SPL.
Assumption: playing 2 (8 ohm) drivers right next to each other will net you 3dB from the extra cone area from the 2 drivers, and the output will also net you 3dB for a final gross of 6dB (if both drivers are driven as their intended load as 8 ohm loads). If it's only 3dB, than I go back to the drawing board.
Each woofer's sensitivity is 87.5dB at 1w/1m.
Here's what I want to do to achieve roughly 87dB of SPL from the woofer pair:
I will wire up the woofer drivers in parallel. I will use an effective load of 7 ohms for each driver due to their impedance curves from the manufacturer.
When parallelled, the woofers give me a 3.5 ohm load.
This is what I propose to do: I will wire up a 3.5 ohm resistor in series with the 2 parallelled woofers.
Here's why:
I calculate that current needed to produce 1W of power for each woofer is the square root of 1/7, which is 0.378A.
This comes from the power equation:
P = (I^2)x(R)
(i.e., Power equals the square of the current times the resistance)
This is the normal amount of current if I would be using just 1 woofer driver and this will give me the 87.5dB. But since I have to use 2 woofers (and not cause a strain on my amp by keeping to a nominal 8 ohm load if I can help it), I want to find a way to still get the 87.5dB SPL with 2 woofers
So, if I put the 3.5 ohm resistor in series with the 2 parallelled woofer, this means the current going to the woofers is divided in 2, for a value of 0.189A for each woofer. The power into each woofer is the square of the current times the resistance: (0.189)^2x7 = 0.25W.
Using the 0.25W power into the woofer, I would get a 6dB drop in SPL for a new SPL of 81.5dB for each woofer.
If I sum the 2 woofer's SPL, I add 3dB when I double the SPL in dB for a value of 84.5dB.
Now, if I can assume that I also get another 3dB for having the 2x increase in the cone area of using 2 woofer, then I add another 3dB to 84.5dB and get 87.5dB, which is where I want to be.
Did I do the math correctly, or am I stuck with a sensitivity of 84.5dB when I series the 3.5 ohm resistor and the 2 parallelled woofers if my input power is 1W/1m?
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2. ### Robert A Stunt Coordinator

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Looks pretty good. Yes you will get 6db off the two woofers if the are powered from the same source. Be careful with your attenuation circuits. If you use resistors instead of an l-pad configuration you may have to compensate with different inductor/capacitor values to still get your desired crossover point. Remember in your calculations that, while resitor circuits and power can be calculated with Ohm's law(V = RI) and Watt's law(P = VI), they are constant, however, inductors, capacitors and speakers themselves are frequency dependent and the resitive nature of each is always changing. Using these equations with R = the impedence is ok, as long as you are only calculating for when the frequencies corralate to that impedence. This is why L-pads are so much better for driver attenuation. As long as you have measured the drivers themselves correctly you should be good to go. Just watch out for a possible deviance in crossover points.
-Rob
P.S. The box looks cool.

3. ### Patrick Sun Moderator Moderator

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Robert, welcome to the HTF, glad you were enticed to post here! Glad you liked the box. I'm still coughing up sawdust...
I think I will forego putting the resistor in series with a set of parallelled woofers due the reasons you brought up. A friend of mine who works in the speaker industry suggested I go with connecting the 2 woofers in series (doubling the impedance) because it would not affect the frequency response, and would maintain the sensitivity, which is the target number I need anyhow (around 87.5dB SPL). Sure, it'll lower the current being sent to each woofer, but in the end, it should work out the same as just having one woofer connected to the crossover, and requires no padding.
But with different load, the crossover does have to be re-adjusted to account for the increase in impedance from the 2 woofers. I have capacitors on order to test out 2 scenarios: woofers in parallel, and woofers in series. (I have plenty of inductor to mix and match with to get what I need in inductance for the crossover.
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4. ### Chris Hoppe Stunt Coordinator

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Looks like you already figured this out, but you generally want to avoid padding down a woofer with a resistor. In your example, the resistor would be sucking as much power as the two woofers!
Also, putting a resistor in series with a woofer would destroy your damping factor, which would make your bass sound flabby, warm and fuzzy. (like a tube amp in fact. )
If your amp can handle it, a good way is to just parallel the woofers and pad down the mid and tweet. This would give a 4 ohm load, and if your amp is OK with this, it would give you a lot more power.

5. ### Patrick Sun Moderator Moderator

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Chris, the problem is that if I parallel the woofers, my SPL goes up to around 93dB while my unpadded tweeter is around 89dB and my midrange is around 90dB unpadded, and I'd need to pad the woofer to bring down their SPLs. Soo...that's why I'm leaning towards doing the woofer in series.
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6. ### Greg Monfort Supporting Actor

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>Did I do the math correctly, or am I stuck with a sensitivity of 84.5dB when I series the 3.5 ohm resistor and the 2 parallelled woofers if my input power is 1W/1m?
====
Sensitivity is how much output for a given voltage signal. Expressed as xxxdB/2.83V.
Efficiency is turning electrical power into mechanical/acoustic power. Expressed as W/m.
I think in terms of voltage gain, rather than current, since solid state amplifiers act as constant voltage sources, and anyway it's what drives the speaker.
E = W*R^0.5, or 1W = 2.83V, so: 87.5dB/1W/m = 87.5dB/2.83V
Wiring drivers in parallel cuts R in half to 4ohms, or 1W = 2V, so:
dB = 20*log(E/2.83) = -3.01dB, or ~84.5dB/2.83V
If you put a resistor in series that matches the paralleled driver's nominal impedance, then sensitivity increases +3dB, so now you're back to ~87.5dB/2.83V (1W).
As long as the drivers are

7. ### Patrick Sun Moderator Moderator

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You sort of lost me here:

8. ### Greg Monfort Supporting Actor

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Because it's bringing it back to the 2.83v (1W/8ohm) reference. dB = 20*log(2.83/2.83) = 0dB, or the same as one.
BTW, how did you highlight that?
GM
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9. ### Patrick Sun Moderator Moderator

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10. ### Greg Monfort Supporting Actor

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Thanks!
GM
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11. ### Mark Seaton Supporting Actor

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Patrick,
It sounds like you are handicapping your center channel's efficiency just to keep the same crossover components as the main speakers. Plain and simple, the efficiency of the system should be ~ that of the least efficient component. Since you mention that the mid and tweet have current divider networks to bring sensitivity down to ~87dB, I would then presume that in your main speakers, the 10" woofer is the limiting factor in the efficiency of the system. So you realy should determine what the actual efficiency of the mid and tweet are, and then see if the paralleled woofers are capable of matching it.
Unless your amplifier is not up to the task, you should have no worry about paralleling the woofers. Also, there will be some amount of loss in output at lower frequencies from what is often reffered to as the "baffle step." Only difference is that in this case the baffle is nearly the size of your RPTV. Once you get the drivers in the box, take some measurments of the raw driver's output levels, and see how things add up.
earlier it sounded like you were trying to keep the impedance of all the drivers to a nominal 8 Ohms. Don't sweat it. As long as each network is scaled for the proper impedance (which is far from constant anyway) you will be fine, and you could gain a bit of headroom for your center channel by making it more efficient.
Mark Seaton

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